class Solution:
  def longestCommonPrefix(self, strs):
    """
    :type strs: List[str]
    :rtype: str
    """
    olds = ''
    rs=''
    if not strs:
      return ''
    if len(strs)==1:
      return strs[0]
    over = False
    print("strs_len:", len(strs))
    i=0 # 單個字符的索引
    while True:
      j=0 # strs中字符串的索引
      while j < len(strs)-1:
        if not strs[j] or not strs[j+1]:# 若是空串
          break
        #print("s1:",len(strs[j]), " s2:", len(strs[j+1]))
        if i >= len(strs[j]): # 如果一個子串比較完了，則結(jié)束循環(huán)
          over = True
          break
        elif i >= len(strs[j+1]):
          j+=1
          over = True
          break
        else:
          over = False
        #print("j:",j,"i: ",i)
        if strs[j][i] != strs[j+1][i]:
          break
        j+=1
      #print("j2:",j)
      if not over and j!=(len(strs)-1): # 如果沒有比較到最后，說明前面出現(xiàn)了不相等的子串
        break
      #print("j:",j,"i: ",i)
      if over: # 如果一個子串已經(jīng)比較結(jié)束，結(jié)束循環(huán)
        rs=strs[j]
        break
      i+=1
      rs=strs[0][:i]
      print("rs:",rs)
    return rs

os.path.commonprefix(list)